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Suppose \(X\) has an exponential distribution with a mean of \(10 .\) Determine the following. (a) \(P(X<5)\) (b) \(P(X<15 \mid X>10)\) (c) Compare the results in parts (a) and (b) and comment on the role of the lack of memory property.

Short Answer

Expert verified
(a) 0.3935; (b) 0.3935; (c) This illustrates the lack of memory property.

Step by step solution

01

Identify Parameters of Exponential Distribution

The exponential distribution is defined by its rate parameter \( \lambda \). Given the mean is 10, we use the relationship \( \text{mean} = \frac{1}{\lambda} \) to find \( \lambda = \frac{1}{10} \).
02

Compute P(X

The cumulative distribution function (CDF) for an exponential distribution is given by \( F(x) = 1 - e^{-\lambda x} \). To find \( P(X < 5) \), plug in \( x = 5 \) and \( \lambda = \frac{1}{10} \): \[F(5) = 1 - e^{-\frac{1}{10} \times 5} = 1 - e^{-0.5} \approx 0.3935.\]
03

Apply Lack of Memory Property for P(X10)

The lack of memory property of the exponential distribution states that \( P(X < a + b \mid X > b) = P(X < a) \). Here, \( a = 5 \) (since 15 > 10 by 5 units), therefore:\[ P(X < 15 \mid X > 10) = P(X < 5) = 1 - e^{-0.5} \approx 0.3935. \]
04

Compare the Results and Comment

The results from steps 2 and 3 are the same, which demonstrates the lack of memory property in the exponential distribution. This property means the probability of the event depends only on the difference \( a \), not on what has occurred prior.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean of Exponential Distribution
The mean or expected value of an exponential distribution is a fundamental parameter. It indicates the average waiting time between occurrences of events, assuming these events occur continuously and independently at a constant average rate. In exponential terms, the mean is expressed as the reciprocal of the rate parameter \( \lambda \), such that \( \text{mean} = \frac{1}{\lambda} \).
For example, if an exponential distribution has a mean of 10, the interpretation is that the average duration (or time interval) between consecutive events or occurrences is 10 units of time.
  • To calculate \( \lambda \), you would take \( \lambda = \frac{1}{\text{mean}} = \frac{1}{10} = 0.1 \). This tells us that events occur at a rate of 0.1 per time unit.
  • Every exponential distribution reflects this reciprocal relationship, meaning that as the mean increases, the rate \( \lambda \) decreases, and vice versa.
Understanding the mean helps in predicting the likeliness of how soon or late an event may occur, which is an essential aspect of the exponential distribution.
Lack of Memory Property
The lack of memory property is a fascinating characteristic of the exponential distribution. This property implies that the distribution's probability for the occurrence of an event in the future is independent of any knowledge of the past. In simpler terms, the probability that you have to wait an additional time \( a \) given you've already waited \( b \) time units is exactly the same as if you hadn't been waiting at all.
Mathematically, this is expressed as:
\[ P(X < a + b \mid X > b) = P(X < a) \]
  • This equation shows the memoryless nature, indicating that future probabilities do not depend on the past elapsed time.
  • It's a unique property that differs from many other probability distributions, making exponential distributions ideal for modeling certain random processes like radioactive decay or various kinds of reliability tests.
So, if you're interested in the time between independent events in a Poisson process, understanding the lack of memory property provides insights into why future probabilities remain consistent regardless of past outcomes.
Cumulative Distribution Function
The cumulative distribution function (CDF) of an exponential distribution provides valuable insights into the probability that a random variable \( X \) will take a value less than or equal to \( x \). It helps in understanding how probabilities accumulate with increasing time.
The CDF function is expressed as:
\[ F(x) = 1 - e^{-\lambda x} \]
  • Here, \( \lambda \) is the rate parameter. This formula calculates the probability that \( X \) falls within a certain range, up to the point \( x \).
  • If you need to find the probability of a specific outcome, you can use this function to compute it directly. For example, if \( \lambda = 0.1 \) and you want to find \( P(X < 5) \), you would solve: \( 1 - e^{-0.1 \times 5} = 1 - e^{-0.5} \).
Employing this tool allows for quickly assessing time-related probabilities crucial in industries dependent on time efficiency. Practically, it models scenarios like service times in queues, time until failure of components, and much more.

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Most popular questions from this chapter

Cholesterol is a fatty substance that is an important part of the outer lining (membrane) of cells in the body of animals. Its normal range for an adult is \(120-240 \mathrm{mg} / \mathrm{dl}\). The Food and Nutrition Institute of the Philippines found that the total cholesterol level for Filipino adults has a mean of \(159.2 \mathrm{mg} / \mathrm{dl}\) and \(84.1 \%\) of adults have a cholesterol level below \(200 \mathrm{mg} / \mathrm{dl}\) (http://www. fnri.dost.gov.ph/). Suppose that the total cholesterol level is normally distributed. (a) Determine the standard deviation of this distribution. (b) What are the quartiles (the \(25 \%\) and \(75 \%\) values) of this distribution? (c) What is the value of the cholesterol level that exceeds \(90 \%\) of the population? (d) An adult is at moderate risk if cholesterol level is more than one but less than two standard deviations above the mean. What percentage of the population is at moderate risk according to this criterion? (e) An adult is thought to be at high risk if his cholesterol level is more than two standard deviations above the mean. What percentage of the population is at high risk? (f) An adult has low risk if cholesterol level is one standard deviation or more below the mean. What percentage of the population is at low risk?

The average height of a woman aged \(20-74\) years is 64 inches in 2002 , with an increase of approximately one inch from 1960 (http://usgovinfo.about,com/od/healthcare). Suppose the height of a woman is normally distributed with a standard deviation of 2 inches. (a) What is the probability that a randomly selected woman in this population is between 58 inches and 70 inches? (b) What are the quartiles of this distribution? (c) Determine the height that is symmetric about the mean that includes \(90 \%\) of this population. (d) What is the probability that five women selected at random from this population all exceed 68 inches?

Suppose that \(X\) has a lognormal distribution with parameters \(\theta=5\) and \(\omega^{2}=9\). Determine the following: (a) \(P(X<13,300)\) (b) The value for \(x\) such that \(P(X \leq x)=0.95\) (c) The mean and variance of \(X\)

The thickness of a laminated covering for a wood surface is normally distributed with a mean of 5 millimeters and a standard deviation of 0.2 millimeter. (a) What is the probability that a covering thickness is greater than 5.5 millimeters? (b) If the specifications require the thickness to be between 4.5 and 5.5 millimeters, what proportion of coverings do not meet specifications?

In an accelerator center, an experiment needs a 1.41 -cmthick aluminum cylinder (http://puhepl.princeton.edu/mumu/ target/Solenoid_Coil.pdf). Suppose that the thickness of a cylinder has a normal distribution with a mean of \(1,41 \mathrm{~cm}\) and a standard deviation of \(0.01 \mathrm{~cm}\). (a) What is the probability that a thickness is greater than \(1.42 \mathrm{~cm} ?\) (b) What thickness is exceeded by \(95 \%\) of the samples? (c) If the specifications require that the thickness is between \(1.39 \mathrm{~cm}\) and \(1.43 \mathrm{~cm}\), what proportion of the samples meet specifications?

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